Brain Teaser No 1

Here goes.

Assume you are floating in space at the "zero gravity" point between the Earth and the Moon. You are given an infinitesimally small nudge towards the Moon and continue in an free fall to the lunar surface. How fast are you travelling, relative to the surface of the Moon, when you impact the Moon?

Assume that the Moon is in its closest position to the Earth when this very likely exercise takes place.

Oh, and I don't have the answer. I'm using the Brain Trust here to get one

- GregB

Insert Witty Signature Line Here

quote:

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Originally posted by Don Atkinson:
Old Martin Payne could act as referee.

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Who are you calling old? I'll admit, though, that I'm not quite as young as my profile,

Mind you, all that power!!

quote:

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and to give Martin a sporting chance as referee

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OK, now you're forgiven.

BTW, agree, Vuk's best chance is to miss, although with 1/3 hitrate I hope he doesn't hit me instead.

cheers, Martin

I found a really neat page on the Wolfram site about 'Astroids' (find it with Google). But the equation of an Astroid only helps with the case where there is only one 'd'. ( x^2/3 + y^2/3 = L^2/3)

Anyway, there is a conspicuous lack of 'elegant' solutions that don't involve equations with square roots in them.

And I can't find one by myself....

So I withdraw!

Paul

We possibly saw a photo of your dad recently, stood between your old Isobariks!

and to give Martin a sporting chance as referee....... with 1/3 hitrate I hope he doesn't hit me instead. Seconded.

But I don't hold out much hope for Mick Parry, who, being the gent he is, only stepped in to calm things down!!

Cheers

Don

You haven't withdrawn, you're only resting

Let me encourage you with a couple of silly psudo-solutions, obtained using Duncan F's infalable 'solution by extensive writing down'

d = 1/L + 1/(L^2) works very well for all values of L>10 ie when the ladder is 10 times longer than the crate!

d = 1/L + 1/(L^2) + L^2.2022/(L^L) which also works very well for L>10 but also gives d = 1 (to a lot of decimal places) when L = sqrt 8 (ie the minimum ratio of crate to ladder length.

Silly! and Bam must be laughing his socks off!)

Cheers

Don

Let's start with the simple statement a = b

Multiply both sides by a
a^2 = ab

Add a^2 - 2ab to both sides
a^2 + a^2 - 2ab = ab + a^2 - 2ab

Simplify
2(a^2 - ab) = a^2 - ab

Finally divide both sides by a^2 - ab
2 = 1

Somewhere there is a disastrous error!

Cheers

Don

quote:

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Finally divide both sides by a^2 - ab

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All of these easily accessible mathematical paradoxes seem to somewhere involve a thinly veiled division by zero.
If a=b.. then a^2=ab
so a^2-ab= 0.. which means that using this entity as a divisor is a guarantee of a 'paradox'.

Ron
Dum spiro audio
Dum audio vivo

All of these easily accessible mathematical paradoxes seem to somewhere involve a thinly veiled division by zero.

Well spotted! whether you spotted the division by zero, or found the solution in one of those easily accessible mathematical paradox books!!

My guess is that most, if not all, of the puzzles in this forum post were first read by the author in some book or other at some stage. I have forgotten many of my sources, but none of mine are original to me! This last one came from Page 341 of the book Fermat's Last Theorem by Simon Singh. Trouble is, if I acknowlege this when posting the puzzle, the answer is usually on page 342!

Getting the answer nevertheless still involves decision making! ie should I try to solve it by THINKING or would it be quicker to run down to the library hoping to get a lucky break and FINDING the solution in a book of paradoxes/puzzles/brainteasers??

Decisions, decisions!!!!

Cheers

Don

I will need time to check this one out (Ron T please note, this means going to the library) but I kind of recall something like

e^(pi*i)=1 or
e^(pi*i)-1=0

close?? ( of course, now I shall sit up all night trying all permutations of these unique numbers until something clicks!!)

Cheers

Don

[This message was edited by Don Atkinson on SUNDAY 27 January 2002 at 22:38.]

I think this was a discovery of Euler, a simple proof is,

e^x = 1 + x + x^2/2! + x^3/3! ...

sin (x) = x - x^3/3! + x^5/5! ...
cos (x) = 1 - x^2/2! + x^4/4! - x^6/6! ...

e^(i*pi) = 1 + i*pi + (i*pi)^2/2!....
= 1 - pi^2/2! + pi^4/4! ... +
i(pi - pi^3/3! + pi^5/5!....)
= cos (pi) + isin(pi)

since cos (pi) = -1 and sin (pi) = 0,

e^(i*pi) = -1

etc... I'm sure there are others. I ended up inside those infinite series for e/sin/cos when under ladders, so it's appropriate.

Now, BAM, give us a break!

Paul

I have now refreshed my memory about binomial expansions so can offer the following

[(L^2 + 1)^0.5] - 1 = L[1 + (1/L^2)^0.5] - 1

= L{1 + [1/2(L^2)] - [1/8(L^4)] + [1/16(L^6)] - [5/128(L^8)] ……} - 1

or in words and letters…..= L{1 + one divided by (2 times (L squared)) - one divided by (8 times (L to the power 4)) + ….} minus 1

This can be simplified to

L - 1 + [1/(2L)] - [1/8(L^3)] + [1/16(L^5)] - [5/128(L^7)].....

This gives a very good approximation using only the first 3 or 4 terms of the expansion, even when L is quite small……much better than my last approximation!

OTOH, I have drawn a dozen pictures to try and find some insight towards an elegant solution, all without finding any inspiration. My note book is full of circles, triangles, squares and bloody ladders! Some of them looking like tessellations.

FWIW the last one looks like two squares, one concentrically (?) inside the other. The outer square is side (d + 1/d + 2) and the inner square is side (d + 1/d). The difference in size is 2m so that there is a 1m wide 'corridor' between the squares. A THIRD square, of side L (=length of ladder) lies 'skew-whiff' BETWEEN the first two squares, touching the inner square at its corners and the outer square at (1 + d) from each corner.

IOW its just four of the original ladders fitted around two walls, a floor and ceiling, of a cubic room!!!

Not much progress huh!! Any body else got any ideas?

Cheers

Don

[This message was edited by Don Atkinson on SUNDAY 03 February 2002 at 22:34.]

The ages of my father my son and myself total 85 years. My father is just twice my age, and the units figure in his age is equal to the age of my son. How old am I?

Cheers

Don

then:

G + F + S = 85 (given)

2F = G (given)

G, like any decimal, can be expressed as 10x+S (given that S is the units of my fathers age)

0 le x le 9 (le means less than or equal to)

and

0 le S le 9

by some algebra, we get a tighter bound that

4 le x le 9

and 6x + S = 34

we than examine some plausible integer combinations of x and S that satisfy this last equation and we get

S = 4

x = 5

and therefore:

G = 54
F = 27
S = 4

voila... (don, you, you fusspot you )

enjoy

ken

[This message was edited by ken c on MONDAY 04 February 2002 at 00:22.]

If you know the length of the ladder the problem gets a lot simpler.

But must remain an exercise for the reader until later....

Paul

Now, my French wasn't that good at school and has steadily declined ever since, however, you might just like to have another crack at the 'voalla' bit on the end!!!!!unless of course its some other language to do with pure maths....

Cheers

Don

lets assume its 3m long tonight, and 4m long tomorrow night and 5.4321m long on Tuesday night and see whether by Wednesday night we've cracked it!!!!

I'm not too sure how long it will be on Thursday night...lets just say L (or should that be HELL) and see whether Sunday to Wednesday has helped.

I'm running out of ideas at the moment and have been reduced to asking my youngest daughter, her boyfriend and one of my nephews (all are doing 1st year sixth form A level maths) for ideas. They have even fewer than me but I'm hoping their teachers can help!!!

But must remain an exercise for the reader until later does this mean you have had a breakthrough Paul???

How many rungs up this ladder do you reckon we've managed so far? recalling the condition of the ladder described by Bam every rung I step on seems to have snapped as soon as I put any weight on it!

And so to bed....

Cheers

Don

me? sulk for the rest of the week i think ...
your fault

enjoy

ken

quote:

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does this mean you have had a breakthrough Paul???

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No.

I don't see how one can get away from the solution to the problem, which is a pig ugly equation, and still produce exactness.

I've found material on this problem from professional mathematicians, learned a little about garage doors and astroid curves, spent rather too long covering paper with badly drawn diagrams of right angles and semi-circles. No elegant exact general answer.

I would solve it this way.

Ladder length 2l, crate 1 cube, bottom of ladder to crate is d, top of ladder to top of crate is 1/d. Crate touches ladder l-x from bottom, l+x from top.

From Pythagoras,

d² = (l - x)² - 1 (1)
1/d² = (l + x)² - 1

So

d²/d² = 1 = ((l - x)² - 1)((l + x)² - 1)

(l-x)²(l+x)² - (l-x)² - (l+x)² + 1 = 1
(l²-x²)² - 2(l² + x²) = 0

x4 - 2(l²-1)x² + l4 -2l² = 0 (2)

which is nicely quadratic in x². At this point I could produce an expression for x² in l and plug it into (1) above and get (hopefully) a familiar equation. However I would rather measure the damn ladder, put the result into (2) and get the actual value of 'x' for this ladder today, then calculate the associated value of 'd' from (1).

I would obviously have a small pocket calculator with a square root function...

Paul

you may receive the mail twice, with .doc attachment -- use whichever works.

enjoy

ken

pythagoras leads to: (L^2)= (1+1/d)^2 + (1+d)^2

which can be recast (after some algebra) to:

L^2 = (d + 1/d)^2 + 2 * (d + 1/d).......(1)

now let x = d + 1/d

solving equation (1) for x, we have:

x = -1 + sqrt(1 + L^2)

now solve for <d> in x = d + 1/d

we get:

d = 0.5 * (x + sqrt(x^2 - 4)) or

d = 0.5 * (x - sqrt(x^2 - 4))

where x = -1 + sqrt(1 + L^2)

i havent read the other emails on this problem. apologies if this has been suggested already.

not sure how "elegant" this is, but, hey, what the hell...

enjoy

ken

I (the emphasis is on the 'I'), I think your solution is very elegant in deed, because...

i havent read the other emails on this problem. apologies if this has been suggested already.

.....the solution that I posted on page 16 (9th entry down) is the same. You will need to wear special glasses to read the attachment but it is just legible.

Of course Bam wasn't too impressed and only gave it 5 out of 10!!!!

I then combined the two basic quadratics together and produced a rather clumsy, single equation for which I got 8/10!!

I am just about to create an attachment showing a totally sophisticated and rather elegant development of that single equation, for which Bam will award me eight and a half out of ten and tell us all we are looking in the wrong direction to find HIS really elegant solution.

However, i think you did rather well ken, and if the scoring was down to me, you would have got nine and a half by now!

Cheers

Don

oh goodie!

"...and tell us all we are looking in the wrong direction to find HIS really elegant solution."

Probably

I'm just glad you are all learning so much about maths and astrophysics and so on as a result of this teaser. This is good, right? I'm hoping these fringe benefits will reduce my chances of being linched if my elegant solution is not elegant enough!

Paul wrote:
"I'm running out of ideas at the moment and have been reduced to asking my youngest daughter, her boyfriend and one of my nephews (all are doing 1st year sixth form A level maths) for ideas. They have even fewer than me but I'm hoping their teachers can help!!!"

Good strategy. I doubt the kids will know but some teachers might. Go for the head of the maths dept, forget the other teachers.

This reminds me of when I was first set this puzzle. I was only about 9 years old and my oncle posed it to me in a rather casual way. Sort of - "here's a little maths puzzle for you Brian, when you have solved it you will be good at maths". Remember the film Kung Fu? It was a little like when his blind master told him it would be time for him to leave when he succeeded in grabbing the marbles out of his hand. It haunted me for 8 years before I gave up and asked my Further Maths teacher. He was an understated mathematician from Cambridge by name of Mr Walker. One day after the lesson I drew a diagram on the whiteboard, showed him my reams of incomplete equations and asked his opinion. He stared at it for just a few moments, quietly muttering words like "interesting" and "yes" and "hmm". He then solved it in just minute without any hesitation, deviation or repetition. When he had done he simply said: "there". He then looked at me with a face that seemed to say "so, what was all the fuss about?". I was distraut. I knew I was in the presence of a God. I wonder where he is now?

BAM
(I've already given you a clue Don even though I said I wouldn't - did you notice it?)

[This message was edited by bam on MONDAY 04 February 2002 at 22:23.]

Its 'kinda neat' because the {-1 + sqrt(Lsq + 1)} part, repeats itself both inside and outside the main sqrt bit.

Although it LOOKS complicated, it is dead easy to use. You only need to calculate the {-1 + sqrt(Lsq + 1)} part once and use it twice, etc and 'Bob's your uncle!'....Voila! (as ken c would say!)

For example if L=(2*sqrt of 2) then the {-1 + sqrt(Lsq + 1)} part = 2 and d=0.5*(2-sqrt0)=1

If L=10 the {-1 + sqrt(Lsq + 1)} part = 9.05 and d=0.11188

Now this is neat, not elegant but neat. Worth eight and a half IMHO.!!

BTW, in the attachment, the '1/2' bit, after the [L^2 + 1] bits, represents 'to the power of 0.5', ie sqrt!! This happens twice, of course.

hope this makes sense.

Cheers

Don

The only clue I recall is on page 16, penultimate post, where you say To win the champagne you need to use less conventional maths.

I don't recall learning less conventional maths!

However, in order to find even this hint (I must have missed the other one!), I had to re-read my postings, and notice how we are going round in ever decreasing circles with rambling variations on the same theme. My above 'neat' soultion was hovering around in late December!!!

Lets hope March doesn't come too soon.

BTW, the e-mail address of Mr Walker would be most appreciated!!

Cheers

Don